# How do you differentiate y=tanx/(2x^3)?

Dec 12, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{x}^{2} {\sec}^{2} x - 3 \tan x}{2 {x}^{4}}$

#### Explanation:

If you are studying maths, then you should learn the Quotient Rule for Differentiation, and practice how to use it:

$\frac{d}{\mathrm{dx}} \left(\frac{u}{v}\right) = \frac{v \frac{\mathrm{du}}{\mathrm{dx}} - u \frac{\mathrm{dv}}{\mathrm{dx}}}{v} ^ 2$, or less formally, $\left(\frac{u}{v}\right) ' = \frac{v \left(\mathrm{du}\right) - u \left(\mathrm{dv}\right)}{v} ^ 2$

I was taught to remember the rule in word; " vdu minus udv all over v squared ". To help with the ordering I was taught to remember the acronym, VDU as in Visual Display Unit.

So with $y = \frac{\tan x}{2 {x}^{3}}$ Then

$\left\{\begin{matrix}\text{Let "u=tanx & => & (du)/dx=sec^2x \\ "And } v = 2 {x}^{3} & \implies & \frac{\mathrm{dv}}{\mathrm{dx}} = 6 {x}^{2}\end{matrix}\right.$

$\therefore \frac{d}{\mathrm{dx}} \left(\frac{u}{v}\right) = \frac{v \frac{\mathrm{du}}{\mathrm{dx}} - u \frac{\mathrm{dv}}{\mathrm{dx}}}{v} ^ 2$
$\therefore \text{ } \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\left(2 {x}^{3}\right) \left({\sec}^{2} x\right) - \left(\tan x\right) \left(6 {x}^{2}\right)}{2 {x}^{3}} ^ 2$
$\therefore \text{ } \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 {x}^{3} {\sec}^{2} x - 6 {x}^{2} \tan x}{4 {x}^{6}}$
$\therefore \text{ } \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{x}^{2} {\sec}^{2} x - 3 \tan x}{2 {x}^{4}}$